Problem: You have found the following ages (in years) of all 4 lions at your local zoo: $ 7,\enspace 10,\enspace 9,\enspace 1$ What is the average age of the lions at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{7 + 10 + 9 + 1}{{4}} = {6.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $0.2$ years $0.04$ years $^2$ $10$ years $3.2$ years $10.24$ years $^2$ $9$ years $2.2$ years $4.84$ years $^2$ $1$ year $-5.8$ years $33.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{0.04} + {10.24} + {4.84} + {33.64}} {{4}} $ $ {\sigma^2} = \dfrac{{48.76}}{{4}} = {12.19\text{ years}^2} $ The average lion at the zoo is 6.8 years old. The population variance is 12.19 years $^2$.